Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.*;
import java.math.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        double[] p = new double[N];
        for (int i = 0; i < N; i++) {
            p[i] = sc.nextDouble();
        }

        Map<Integer, Double> map = new HashMap<>();
        map.put(0, 1.0);
        for (int i = 0; i < N; i++) {
            Map<Integer, Double> tmp = new HashMap<>();
            for (int key : map.keySet()) {
                double last = map.get(key);
                tmp.put(key + 1, tmp.getOrDefault(key + 1, 0.0) + last * (1 - p[i]));
                tmp.put(key + 10000, tmp.getOrDefault(key + 10000, 0.0) + last * p[i]);
            }
            map = tmp;
        }

        double res = 0.0;

        for (int key : map.keySet()) {
            if (key / 10000 > key % 10000) {
                res += map.get(key);
            }
        }

        System.out.println(res);
    }
}